3.29.10 \(\int \frac {121+400 x-1100 x^2+2500 x^4+(220-1000 x^2) (i \pi +\log (-\log (\frac {\log (4)}{2})))+100 (i \pi +\log (-\log (\frac {\log (4)}{2})))^2}{121-1100 x^2+2500 x^4+(220-1000 x^2) (i \pi +\log (-\log (\frac {\log (4)}{2})))+100 (i \pi +\log (-\log (\frac {\log (4)}{2})))^2} \, dx\)

Optimal. Leaf size=38 \[ x-\log (4)+\frac {2}{\frac {1}{2}-25 x^2+5 \left (1+i \pi +\log \left (-\log \left (\frac {\log (4)}{2}\right )\right )\right )} \]

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Rubi [A]  time = 0.14, antiderivative size = 26, normalized size of antiderivative = 0.68, number of steps used = 5, number of rules used = 5, integrand size = 118, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {1994, 28, 1814, 21, 8} \begin {gather*} x+\frac {4}{-50 x^2+10 i \pi +11+10 \log (-\log (\log (2)))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(121 + 400*x - 1100*x^2 + 2500*x^4 + (220 - 1000*x^2)*(I*Pi + Log[-Log[Log[4]/2]]) + 100*(I*Pi + Log[-Log[
Log[4]/2]])^2)/(121 - 1100*x^2 + 2500*x^4 + (220 - 1000*x^2)*(I*Pi + Log[-Log[Log[4]/2]]) + 100*(I*Pi + Log[-L
og[Log[4]/2]])^2),x]

[Out]

x + 4/(11 + (10*I)*Pi - 50*x^2 + 10*Log[-Log[Log[2]]])

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 1814

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[P
olynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[((a
*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*(p + 1)), Int[(a + b*x^2)^(p + 1)*ExpandToS
um[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]

Rule 1994

Int[(Pq_)*(u_)^(p_.), x_Symbol] :> Int[Pq*ExpandToSum[u, x]^p, x] /; FreeQ[p, x] && PolyQ[Pq, x] && TrinomialQ
[u, x] &&  !TrinomialMatchQ[u, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {121+400 x-1100 x^2+2500 x^4+\left (220-1000 x^2\right ) \left (i \pi +\log \left (-\log \left (\frac {\log (4)}{2}\right )\right )\right )+100 \left (i \pi +\log \left (-\log \left (\frac {\log (4)}{2}\right )\right )\right )^2}{2500 x^4-(11 i-10 \pi +10 i \log (-\log (\log (2))))^2-100 x^2 (11+10 i \pi +10 \log (-\log (\log (2))))} \, dx\\ &=2500 \int \frac {121+400 x-1100 x^2+2500 x^4+\left (220-1000 x^2\right ) \left (i \pi +\log \left (-\log \left (\frac {\log (4)}{2}\right )\right )\right )+100 \left (i \pi +\log \left (-\log \left (\frac {\log (4)}{2}\right )\right )\right )^2}{\left (2500 x^2-50 (11+10 i \pi +10 \log (-\log (\log (2))))\right )^2} \, dx\\ &=\frac {4}{11+10 i \pi -50 x^2+10 \log (-\log (\log (2)))}+\frac {25 \int \frac {2 (11 i-10 \pi +10 i \log (-\log (\log (2))))^2+100 x^2 (11+10 i \pi +10 \log (-\log (\log (2))))}{2500 x^2-50 (11+10 i \pi +10 \log (-\log (\log (2))))} \, dx}{11+10 i \pi +10 \log (-\log (\log (2)))}\\ &=\frac {4}{11+10 i \pi -50 x^2+10 \log (-\log (\log (2)))}+\int 1 \, dx\\ &=x+\frac {4}{11+10 i \pi -50 x^2+10 \log (-\log (\log (2)))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 26, normalized size = 0.68 \begin {gather*} x-\frac {4}{-11-10 i \pi +50 x^2-10 \log (-\log (\log (2)))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(121 + 400*x - 1100*x^2 + 2500*x^4 + (220 - 1000*x^2)*(I*Pi + Log[-Log[Log[4]/2]]) + 100*(I*Pi + Log
[-Log[Log[4]/2]])^2)/(121 - 1100*x^2 + 2500*x^4 + (220 - 1000*x^2)*(I*Pi + Log[-Log[Log[4]/2]]) + 100*(I*Pi +
Log[-Log[Log[4]/2]])^2),x]

[Out]

x - 4/(-11 - (10*I)*Pi + 50*x^2 - 10*Log[-Log[Log[2]]])

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fricas [A]  time = 0.55, size = 33, normalized size = 0.87 \begin {gather*} \frac {50 \, x^{3} - 10 \, x \log \left (\log \left (\log \relax (2)\right )\right ) - 11 \, x - 4}{50 \, x^{2} - 10 \, \log \left (\log \left (\log \relax (2)\right )\right ) - 11} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((100*log(log(log(2)))^2+(-1000*x^2+220)*log(log(log(2)))+2500*x^4-1100*x^2+400*x+121)/(100*log(log(l
og(2)))^2+(-1000*x^2+220)*log(log(log(2)))+2500*x^4-1100*x^2+121),x, algorithm="fricas")

[Out]

(50*x^3 - 10*x*log(log(log(2))) - 11*x - 4)/(50*x^2 - 10*log(log(log(2))) - 11)

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giac [A]  time = 0.31, size = 19, normalized size = 0.50 \begin {gather*} x - \frac {4}{50 \, x^{2} - 10 \, \log \left (\log \left (\log \relax (2)\right )\right ) - 11} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((100*log(log(log(2)))^2+(-1000*x^2+220)*log(log(log(2)))+2500*x^4-1100*x^2+400*x+121)/(100*log(log(l
og(2)))^2+(-1000*x^2+220)*log(log(log(2)))+2500*x^4-1100*x^2+121),x, algorithm="giac")

[Out]

x - 4/(50*x^2 - 10*log(log(log(2))) - 11)

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maple [A]  time = 0.54, size = 18, normalized size = 0.47

method	result	size
risch	\(x +\frac {2}{5 \left (-5 x^{2}+\ln \left (\ln \left (\ln \relax (2)\right )\right )+\frac {11}{10}\right )}\)	\(18\)
gosper	\(\frac {-50 x^{3}+10 x \ln \left (\ln \left (\ln \relax (2)\right )\right )+11 x +4}{-50 x^{2}+10 \ln \left (\ln \left (\ln \relax (2)\right )\right )+11}\)	\(34\)
norman	\(\frac {-50 x^{3}+4+\left (10 \ln \left (\ln \left (\ln \relax (2)\right )\right )+11\right ) x}{-50 x^{2}+10 \ln \left (\ln \left (\ln \relax (2)\right )\right )+11}\)	\(34\)
default	\(x +\frac {8800+8000 \ln \left (\ln \left (\ln \relax (2)\right )\right )}{\left (-2000 \ln \left (\ln \left (\ln \relax (2)\right )\right )-2200\right ) \left (50 x^{2}-10 \ln \left (\ln \left (\ln \relax (2)\right )\right )-11\right )}\)	\(38\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((100*ln(ln(ln(2)))^2+(-1000*x^2+220)*ln(ln(ln(2)))+2500*x^4-1100*x^2+400*x+121)/(100*ln(ln(ln(2)))^2+(-100
0*x^2+220)*ln(ln(ln(2)))+2500*x^4-1100*x^2+121),x,method=_RETURNVERBOSE)

[Out]

x+2/5/(-5*x^2+ln(ln(ln(2)))+11/10)

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maxima [A]  time = 0.63, size = 19, normalized size = 0.50 \begin {gather*} x - \frac {4}{50 \, x^{2} - 10 \, \log \left (\log \left (\log \relax (2)\right )\right ) - 11} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((100*log(log(log(2)))^2+(-1000*x^2+220)*log(log(log(2)))+2500*x^4-1100*x^2+400*x+121)/(100*log(log(l
og(2)))^2+(-1000*x^2+220)*log(log(log(2)))+2500*x^4-1100*x^2+121),x, algorithm="maxima")

[Out]

x - 4/(50*x^2 - 10*log(log(log(2))) - 11)

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mupad [B]  time = 0.11, size = 19, normalized size = 0.50 \begin {gather*} x+\frac {4}{-50\,x^2+10\,\ln \left (\ln \left (\ln \relax (2)\right )\right )+11} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((400*x - log(log(log(2)))*(1000*x^2 - 220) + 100*log(log(log(2)))^2 - 1100*x^2 + 2500*x^4 + 121)/(100*log(
log(log(2)))^2 - log(log(log(2)))*(1000*x^2 - 220) - 1100*x^2 + 2500*x^4 + 121),x)

[Out]

x + 4/(10*log(log(log(2))) - 50*x^2 + 11)

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sympy [A]  time = 0.39, size = 24, normalized size = 0.63 \begin {gather*} x - \frac {4}{50 x^{2} - 11 - 10 \log {\left (- \log {\left (\log {\relax (2 )} \right )} \right )} - 10 i \pi } \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((100*ln(ln(ln(2)))**2+(-1000*x**2+220)*ln(ln(ln(2)))+2500*x**4-1100*x**2+400*x+121)/(100*ln(ln(ln(2)
))**2+(-1000*x**2+220)*ln(ln(ln(2)))+2500*x**4-1100*x**2+121),x)

[Out]

x - 4/(50*x**2 - 11 - 10*log(-log(log(2))) - 10*I*pi)

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